Prove that if A is a square matrix then A and A transpose have same Eigen values.
Kindly help me how to prove this by generally not considering matrix by ourselves.
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What does "Differentiation" have to do with this question? – DonAntonio Nov 11 '13 at 13:36
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Do you already know any matrix is similar to its transpose? – DonAntonio Nov 11 '13 at 13:38
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sorry,not by differentiation.its wrote wrongly – Ali khan Nov 11 '13 at 13:39
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I changed the title altogether to make more explicit. – DonAntonio Nov 11 '13 at 13:41
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Now what will be the proof? – Ali khan Nov 11 '13 at 13:43
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If the answer to DonAntonio's comment is "no", do you know that $\det(B)=\det(B^T)$ for any square matrix $B$? – Casteels Nov 11 '13 at 13:43
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dan's answer is what I had in mind. – Casteels Nov 11 '13 at 13:53
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The eigen-values solve $det(A-I\lambda)=0$. But $det(A^{T}-I\lambda)=det((A-I\lambda)^{T})=det((A-I\lambda))$ by virtue of the nature of the computation of the determinant. So we may notice that both $det(A-I\lambda)$ and $det(A^T-I\lambda)$ must have the same solutions. And so the same eigen-values.
dan
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$$A\sim A^t\implies \exists\,\text{invertible}\;P\;\;s.t.\;\;P^{-1}AP=A^t\implies$$
$$\det (xI-A^t)=\det(xI-P^{-1}AP)=\det\left[P^{-1}\left(xI-A\right)P\right]=\det(xI-A)\ldots$$
DonAntonio
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I've no idea what you know and what you don't, @user52045, and you didn't answer to my question in the comments. – DonAntonio Nov 11 '13 at 13:54
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